What is (\partial)/(\partial y)(3/(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(3/(x^2+y^2)) is -(6y)/((x^2+y^2)^2)

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(\partial)/(\partial y)(3/(x^2+y^2))

\frac{\partial}{\partial y} (\frac{3}{x ^{2} + y ^{2}})

- \frac{6 y}{( x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{3}{x ^{2} + y ^{2}})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 3 \frac{\partial}{\partial y} ((x^{2} + y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial y} (x^{2} + y^{2})

= - \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial y} (x^{2} + y^{2})

\frac{\partial}{\partial y} (x^{2} + y^{2}) = 2 y

= 3 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 y)

Simplify

3 (- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 y) : - \frac{6 y}{( x ^{2} + y ^{2} ) ^{2}}

= - \frac{6 y}{( x ^{2} + y ^{2} ) ^{2}}

\frac{d}{d x} (3 x^{3} (x^{2} + 4)^{4})

\int 9 - A^{2} d A

x = y^{2} - 2 y, x - y - 4 = 0

h^{^{'}}

\int cos^{3} (θ) d θ

What is (\partial)/(\partial y)(3/(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(3/(x^2+y^2)) is -(6y)/((x^2+y^2)^2)